Standard entropy of `X_(2)` , `Y_(2)` and `XY_(3)` are `60, 40 ` and `50JK^(-1)mol^(-1)` , respectively. For the reaction, `(1)/(2)X_(2)+(3)/(2)Y_(2)rarrXY_(3),DeltaH=-30KJ` , to be at equilibrium, the temperature will be:

Options

A. `750 K`
B. `1000 K`
C. `1250 K`
D. `500 K`

Answer

A

Explanation

According to thermodynamics,
`Delta_(r) G^(@) = Delta_(r) H^(@) = T Delta_(r) S^(@)`
At equilibrium, `Delta_(r) G^(@) = 0`. Thus,
`= 0 Delta_(r) H^(@) - T Delta S^(@)`
or `T = (Delta_(r) H^(@))/(Delta_(r) S^(@))`
We are given `Delta_(r) H^(@) = - 30 kJ`. To calculate `Delta_(r) S^(@)`, we use the following relationship:
`Delta_(r) S^(@) = sum_(i) a_(i) S_(m)^(@)` (products) `- sum_(i) b_(i) S_(@)` (reactant s)
`= S_(m)^(@) (XY_(3)) - [(1)/(2) S_(m)^(@) (X_(2)) + (3)/(2) S_(m)^(@) (Y_(2))]`
`= (50) - [(1)/(2) (60) + (3)/(2) (40)]`
`= (50) - (30 + 60) = - 40 J K^(-1) mol^(-1)`
Substituting this result in the above equation, we get
`T = (- 30 xx 10^(3) J mol^(-1))/(-40 J K^(-1) mol^(-1)) = 750 K`